Table of Contents
Understanding the Substitution Method
In this chapter, we focus on one specific way to solve a system of linear equations in two variables: the substitution method. The general idea of a system of linear equations and what it means to solve one is assumed from the parent chapter.
The substitution method is based on a simple idea:
- Solve one of the equations for one variable in terms of the other.
- Substitute this expression into the other equation.
- Solve the resulting single-variable equation.
- Use that value to find the other variable.
- (If needed) Check your solution in the original equations.
The steps are always similar; what changes is how convenient each step is for a given system.
Step-by-Step Procedure
Consider a system in two variables $x$ and $y$:
$$
\begin{cases}
\text{Equation (1)} \\
\text{Equation (2)}
\end{cases}
$$
A standard procedure for substitution:
- Choose one equation and one variable
Pick the equation where a variable is easiest to isolate (for example, if it already appears with coefficient $ or $-1$). - Solve that equation for the chosen variable
Rearrange the equation so that one variable is alone on one side.
For example, from
$$
2x + y = 7
$$
you might solve for $y$:
$$
y = 7 - 2x.
$$ - Substitute into the other equation
Wherever that variable appears in the other equation, replace it with the expression you just found.
For example, if the other equation is
$$
3x - y = 1,
$$
then using $y = 7 - 2x$ gives
$$
3x - (7 - 2x) = 1.
$$ - Solve the resulting one-variable equation
Now there is only one variable (here, $x$). Solve it as a regular linear equation. - Back-substitute to find the other variable
Take the value you found and plug it back into the expression from step 2 (or into either original equation) to find the second variable. - (Optional but recommended) Check
Substitute both values into both original equations to verify they satisfy the system.
Worked Example: One Variable Easily Isolated
Solve the system:
$$
\begin{cases}
y = 3x + 1 \\
2x + y = 10
\end{cases}
$$
- Choose an equation and variable
The first equation already has $y$ isolated: $y = 3x + 1$. - Substitute into the other equation
In x + y = 10$, replace $y$ with x + 1$:
$$
2x + (3x + 1) = 10.
$$ - Solve for $x$
$$
2x + 3x + 1 = 10 \
5x + 1 = 10 \
5x = 9 \
x = \frac{9}{5}.
$$ - Find $y$
Use $y = 3x + 1$:
$$
y = 3\left(\frac{9}{5}\right) + 1 = \frac{27}{5} + 1 = \frac{27}{5} + \frac{5}{5} = \frac{32}{5}.
$$
So the solution is
$$
(x, y) = \left(\frac{9}{5}, \frac{32}{5}\right).
$$
Example: Isolating a Variable First
Solve:
$$
\begin{cases}
2x + 3y = 7 \\
x - y = 1
\end{cases}
$$
- Choose an equation and variable
The second equation is easy to solve for $x$ or $y$. Let’s solve for $x$:
$$
x - y = 1 \quad \Rightarrow \quad x = 1 + y.
$$ - Substitute into the first equation
In x + 3y = 7$, replace $x$ with + y$:
$$
2(1 + y) + 3y = 7.
$$ - Solve for $y$
$$
2 + 2y + 3y = 7 \
2 + 5y = 7 \
5y = 5 \
y = 1.
$$ - Find $x$
Using $x = 1 + y$:
$$
x = 1 + 1 = 2.
$$
So the solution is $(x, y) = (2, 1)$.
Choosing Which Variable to Substitute
You usually have a choice: solve for $x$ or for $y$, and from either equation. Some choices make the arithmetic simpler.
Guidelines:
- If a variable already has coefficient $1$ or $-1$, solve for that variable.
- Avoid introducing fractions early if you can.
- If both equations are already in forms like $y = \dots$ or $x = \dots$, substitution is especially convenient.
Example system:
$$
\begin{cases}
4x - y = 5 \\
3x + 2y = 1
\end{cases}
$$
Here, $-y$ has coefficient $-1$. Solving $4x - y = 5$ for $y$ avoids fractions:
$$
4x - y = 5 \Rightarrow -y = 5 - 4x \Rightarrow y = 4x - 5.
$$
Then substitute $y = 4x - 5$ into the second equation.
Handling Fractions in Substitution
Sometimes fractions are unavoidable. You should still follow the same steps; just handle the arithmetic carefully.
Solve:
$$
\begin{cases}
2x + y = 1 \\
x - 2y = 4
\end{cases}
$$
- Solve for $y$ from the first equation
$$
2x + y = 1 \Rightarrow y = 1 - 2x.
$$ - Substitute into the second equation
$$
x - 2(1 - 2x) = 4.
$$ - Solve for $x$
$$
x - 2 + 4x = 4 \
5x - 2 = 4 \
5x = 6 \
x = \frac{6}{5}.
$$ - Find $y$
$$
y = 1 - 2\left(\frac{6}{5}\right) = 1 - \frac{12}{5} = \frac{5}{5} - \frac{12}{5} = -\frac{7}{5}.
$$
Solution:
$$
(x, y) = \left(\frac{6}{5}, -\frac{7}{5}\right).
$$
If you end up with fractions early, you can sometimes clear them by multiplying the entire equation by the denominator before substituting, to keep later steps simpler.
Recognizing Special Cases with Substitution
Substitution is also useful to recognize when a system has no solution or infinitely many solutions.
After substituting and simplifying, you may get:
- A false statement, like $0 = 5$.
This means the system is inconsistent: the lines are parallel and never meet, so there is no solution. - A true identity, like $0 = 0$.
This means both equations describe the same line, and there are infinitely many solutions (every point on that line).
No Solution Example
Solve:
$$
\begin{cases}
y = 2x + 1 \\
4x - 2y = 6
\end{cases}
$$
Substitute $y = 2x + 1$ into the second equation:
$$
4x - 2(2x + 1) = 6 \\
4x - 4x - 2 = 6 \\
-2 = 6.
$$
This is impossible, so there is no solution.
Infinitely Many Solutions Example
Solve:
$$
\begin{cases}
y = 3x - 2 \\
6x - 2y = 4
\end{cases}
$$
Substitute $y = 3x - 2$ into the second equation:
$$
6x - 2(3x - 2) = 4 \\
6x - 6x + 4 = 4 \\
4 = 4.
$$
This is always true, so the two equations represent the same line. There are infinitely many solutions (all points on $y = 3x - 2$).
Comparison with the Elimination Method
The elimination method (covered in another chapter) combines equations to eliminate a variable by adding or subtracting them.
Substitution is often preferred when:
- One equation is already solved for a variable (e.g., $y = 5x - 3$).
- A variable has coefficient $1$ or $-1$, making isolation simple.
- You want a more direct use of the idea “replace a variable with an equivalent expression.”
Elimination may be more convenient when both equations are in standard form and coefficients align nicely for canceling a variable.
You should be comfortable recognizing when substitution leads to simpler work and when another method might be better.
Common Mistakes and How to Avoid Them
- Forgetting parentheses during substitution
When you substitute an expression, always use parentheses: - Write $3(2x - 1)$, not $3 \cdot 2x - 1$.
- Write $- (x + 4)$, not $-x + 4$.
- Substituting into the same equation you solved
After you solve one equation for a variable, substitute into the other equation, not back into the same one. - Arithmetic errors
Be careful with: - Combining like terms
- Distributive property
- Signs when subtracting and distributing negatives
- Stopping after finding only one variable
Remember: a solution to the system is an ordered pair $(x, y)$. Always compute both values. - Not recognizing special cases
If you simplify and both variables disappear: - If you get a false statement (e.g., $5 = -1$), conclude no solution.
- If you get a true statement (e.g., $0 = 0$), conclude infinitely many solutions.
Practice Structure (What to Try on Your Own)
When practicing substitution, try problems of these types:
- One equation already solved for a variable:
$$
\begin{cases}
y = -2x + 3 \
5x + y = 7
\end{cases}
$$ - Need to isolate a variable first:
$$
\begin{cases}
3x + 4y = 10 \
x - 2y = -1
\end{cases}
$$ - Systems that lead to fractions:
$$
\begin{cases}
x + y = 2 \
2x - 3y = -4
\end{cases}
$$ - Systems with no solution or infinitely many solutions:
$$
\begin{cases}
y = 4x - 1 \
8x - 2y = 10
\end{cases}
$$
For each system, follow the substitution steps carefully, and then check your answer in both original equations.